Answer :

Meochi
The answer to the question is n^2-4/n^2-3 n ≠±√6±√3

Answer:

The simplified form is [tex]=\frac{(n-2)(n+2)}{(n-\sqrt3)(n+\sqrt3)}[/tex]

The variable n can not be equals to [tex]\pm\sqrt3[/tex] as for these value denominator equals to zero which becomes an indeterminate form.

Step-by-step explanation:

we have to simplify the rational expression and state the restrictions on the variable.

[tex]\frac{n^4-10n^2+24}{n^4-9n^2+18}[/tex]

We can now see that both the numerator and denominator are quadratic trinomials in [tex]n^2.\[/tex]

We split the middle terms as follows;

[tex]\frac{n^4-10n^2+24}{n^4-9n^2+18}=\frac{n^4-6n^2-4n^2+24}{n^4-3n^2-6n^2+18}[/tex]

          [tex]=\frac{n^2(n^2-6)-4(n^2-6)}{n^2(n^2-3)-6(n^2-3)}[/tex]

          [tex]=\frac{(n^2-4)(n^2-6)}{(n^2-6)(n^2-3)}[/tex]

          [tex]=\frac{(n-2)(n+2)}{(n-\sqrt3)(n+\sqrt3)}[/tex]

which is the simplified form of given expression.

The variable n can not be equals to [tex]\pm\sqrt3[/tex] as for these value denominator equals to zero which becomes an indeterminate form.

Other Questions